Are breaks in the continuity of a graph scalar with regards to multiplying a constant?

Question

I think I'm probably right but seeing how I've recently begun to go down the rabbit hole in mathematics that tends to produce all sorts of inconsistencies with any advanced conjectures that I make, I decided to ask if one of my more (self-apparent) opinions was actually true.

In general, will a graphs continuity not change when its corresponding function is multiplied by a constant, and will any and all gaps within a function have their width scaled by a factor of the constant. For instance if a piece wise function had a gap of 1 at point 0, would multiplying by 2 change the gaps width by 2?

If there are times where such a conjecture breaks down, please give me an example (beyond the obvious function-decimating multiplication by 0).

Answer 1

This question was already answered in the comments by Greg Martin; I'm posting an answer because we like to keep our unanswered queue uncluttered. I also want to address a possible extension which you may want that allows "discontinuities of the second kind"; that is, where the limit does not exist. For instance, perhaps you want to say that $f:x\mapsto \sin(1/x)$ has a jump of size 2 at $x=0$? In this case, you are really talking about $\lim\inf$ and $\lim\sup$ rather than one-sided limits.

Side Comments for the OP

It seems that for your program, you are less interested in such things than in continuities of the first kind. But if you do not want to consider this case then feel free to replace everything with one-sided limits; the proof should be identical.

Also, you have used some nonstandard terminology in this post and in others, which has led to some confusion. I would therefore ask that you read following proof for its linguistic as well as its mathematical content.

Let $c>0$, and $g=cf$ be a scaling of the function $f$. If $x$ is a point of discontinuity for $f$, wish to show that $p$ is a point of discontinuity for $g$ as well, with $$\limsup_{x\to p}~ g(x) - \liminf_{x\to p}~ g(x) = c\left(\limsup_{x\to p}~ f(x)-\liminf_{x\to p}~ f(x)\right)$$

Recall that $p$ is point of discontinuity for $g$ if and only if $$\limsup_{x\to p}~ g(x) - \liminf_{x\to p}~ g(x)>0$$ and therefore it suffices to prove the equality, since the LHS is positive if and only if the RHS is.

At this point, where we go in the proof depends a lot on precisely which definition you're using for limsup and liminf. I will use the subsequential limits definition.

Let $\varepsilon>0$ and suppose $x_n\to p$ is a sequence such that $$\limsup_{x\to p}~ f(x)-\lim_{n\to\infty}f(x_n)<\varepsilon/c$$ Multiply through by $c$ to obtain $$c\limsup_{x\to p}~ f(x)-c\lim_{n\to\infty}f(x_n)<\varepsilon.$$

At this point, we could prove that we can move $c$ inside the limit, but I would rather use the opportunity to show how to cite prior arguments, as well as speak somewhat post-rigorously.

Since $\lim$ is a linear operator, we can move $c$ inside the limit symbol. Then $cf=g$, and so $$c\limsup_{x\to p}~ f(x)-\lim_{n\to\infty}g(x_n)<\varepsilon.$$ Since this holds for all $\varepsilon>0$, we have shown that $$\limsup_{x\to p}~ g(x)=c\limsup_{x\to p}~ f(x).$$

An analogous argument works for liminf, and thus factoring out the constant factor of $c$ gives the desired result.

Finally, we need to settle the case of $c<0$. In this case, we need to prove that $$\limsup_{x\to p}~ g(x) - \liminf_{x\to p}~ g(x) = c\left(\liminf_{x\to p}~ f(x)-\limsup_{x\to p}~ f(x)\right),$$ which can be reduced to the previous argument using the identity $\liminf f = -\limsup -f$.

If you are not convinced about this reduction, you should go through the details. Indeed, the reduction is not totally obvious and it's reasonable to think I might be sweeping something under the rug here.

Even less obvious is the also true fact, that this does recover your intuitive description of negative gaps in the case of Type-I discontinuities. Again, if this is not clear you should go through the details. (Admittedly: I would do this one myself, but I am not precisely clear on what from a negative gap, but I am convinced that any reasonable definition you give should follow from this argument.)

Finally, note that this proof goes through even when $x$ is not a point of discontinuity for $f$, or when it is but $c=0$. The reason you lose discontinuity in this case is because of the defining property I cited in the second paragraph: the difference needs to be strictly positive.